3.10-bo'lim : Yashirin farqlash
9. \(y'\) ni \(\tan \left( {{x^2}{y^4}} \right) = 3x + {y^2}\) uchun yashirin farqlash orqali toping.
Maslahat: \(y\) haqiqatan ham \(y\left( x \right)\) ekanligini unutmang va shuning uchun biz \(y\) ishtirokidagi atamalarning hosilasini olishda zanjir qoidasidan foydalanishimiz kerak! Shuni ham unutmangki, \(y\) haqiqatan ham \(y\left( x \right)\) bo'lgani uchun bizda muammoga ko'milgan Mahsulot va/yoki Quotient qoidasi bo'lishi mumkin.
Yechim
Birinchidan, biz \(x\) ga nisbatan hamma narsaning hosilasini olishimiz kerak va biz \(y\) haqiqatan ham \(y\left( x \right)\) ekanligini esga olishimiz kerak, shuning uchun biz \(y\) ishtirokidagi atamalarning hosilasini olishda zanjir qoidasidan foydalanishimiz kerak. Bu shuningdek, chap tomondagi birinchi tangensda zanjir qoidasini bajarayotganda, "ichki atama" ni farqlashda mahsulot qoidasini bajarishimiz kerakligini anglatadi.
\(x\) ga nisbatan farqlash shuni beradi,
\[\left( {2x\,{y^4} + 4{x^2}{y^3}y'} \right){\sec ^2}\left( {{x^2}{y^4}} \right) = 3 + 2y\,y'\]2-qadam
Nihoyat, biz qilishimiz kerak bo'lgan yagona narsa buni \(y'\) uchun hal qilishdir (ba'zi potentsial tartibsiz algebra bilan).
\[\begin{align*}2x\,{y^4}{\sec ^2}\left( {{x^2}{y^4}} \right) + 4{x^2}{y^3}y'{\sec ^2}\left( {{x^2}{y^4}} \right) & = 3 + 2y\,y'\\ \left( {4{x^2}{y^3}{{\sec }^2}\left( {{x^2}{y^4}} \right) - 2y} \right)y' & = 3 - 2x\,{y^4}{\sec ^2}\left( {{x^2}{y^4}} \right)\\ y' & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{3 - 2x\,{y^4}{{\sec }^2}\left( {{x^2}{y^4}} \right)}}{{4{x^2}{y^3}{{\sec }^2}\left( {{x^2}{y^4}} \right) - 2y}}}}\end{align*}\]